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bazaltina [42]
3 years ago
8

An object is moving in a straight line at a constant speed.What will happen to the object if no outside force acts on the object

?
A The speed and direction of the object will change.

B The speed and direction of the object will not change.

C The direction of the object will change; the speed will remain constant.

D The speed of the object will change; the direction will remain constant.

I think it would be C or D?
Chemistry
1 answer:
ruslelena [56]3 years ago
7 0
C is correct answer.
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Five major.

From lowest to highest, the major layers are the troposphere, stratosphere, mesosphere, thermosphere and exosphere. Troposphere.
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How many elements are in the periodic table?
gogolik [260]

There are <u>118</u> elements.

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Chemistry: An oxygen gas container has a volume of 20.0 L. How many grams of oxygen are in the container, if the gas has a press
Morgarella [4.7K]

Answer:

There are 29.4 grams of oxygen in the container

Explanation:

<u>Step 1: </u>Data given

Volume = 20.0 L

Pressure = 845 mmHg

Temperature = 22.0 °C

Molar mass of O2 = 32 g/mol

<u>Step 2:</u> Ideal gas law

p*V = n*R*T

⇒ p = the pressure of the gas = 845 mmHg = 1.11184

⇒ V = the volume of the gas = 20.0 L

⇒ n = the number of moles = TO BE DETERMINED

⇒R = the gasconstant = 0.08206 L*atm/K*mol

⇒ T = the temperature = 22°C + 273 = 295 Kelvin

n = (p*V)/(R*T)

n = (1.11184*20.0)/(0.08206*295)

n = 0.9186 moles

<u>Step 3:</u> Calculate mass of NO2

Mass of O2 = Moles O2 * Molar mass O2

Mass of O2 = 0.9186 moles * 32 g/mol

Mass of O2 = 29.4 grams

There are 29.4 grams of oxygen in the container

8 0
3 years ago
Select the correct answer. Why were industrial and agricultural surpluses a problem for the US economy? A. The United States did
almond37 [142]

Answer:

the correct answers among them is C

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50cm3 of 1 mol/dm3 HCl at 30°C was mixed with 50cm3 of 1mol/dm3 NaOH at 30°C in a styrofoam calorimeter. The temperature of the
trapecia [35]

Answer:

-21 kJ·mol⁻¹  

Explanation:

Data:

                    H₃O⁺ +  OH⁻ ⟶ 2H₂O

       V/mL:    50         50  

c/mol·dm⁻³:   1.0         1.0

     

ΔT = 4.5 °C  

       C = 4.184 J·°C⁻¹g⁻¹

C_cal = 50 J·°C⁻¹

Calculations:

(a) Moles of acid

\text{Moles of acid} = \text{0.050 dm}^{3} \times \dfrac{\text{1.0 mol}}{\text{1 dm}^{3}} = \text{0.050 mol}\\\\\text{Moles of base} = \text{0.050 dm}^{3} \times \dfrac{\text{1.0 mol}}{\text{1 dm}^{3}} = \text{0.050 mol}

So, we have 0.050 mol of reaction

(b) Volume of solution

V = 50 dm³ + 50 dm³ = 100 dm³

(c) Mass of solution

\text{Mass of solution} = \text{100 dm}^{3} \times \dfrac{\text{1.00 g}}{\text{1 dm}^{3}} = \text{100 g}

(d) Calorimetry

There are three energy flows in this reaction.

q₁ = heat from reaction

q₂ = heat to warm the water

q₃ = heat to warm the calorimeter

q₁ + q₂ + q₃ = 0

     nΔH   +         mCΔT       + C_calΔT = 0

0.050ΔH + 100×4.184×4.5 +   50×4.5  = 0

0.050ΔH +          1883        +      225    = 0

                                  0.050ΔH + 2108 = 0

                                              0.050ΔH = -2108

                                                        ΔH = -2108/0.0500

                                                              = -42 000 J/mol

                                                              = -42 kJ/mol

This is the heat of reaction for the formation of 2 mol of water

The heat of reaction for the formation of mol of water is -21 kJ·mol⁻¹.

5 0
3 years ago
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